## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $\rho = 2.34\times 10^{17}~kg/m^3$ (b) $r = 12.7~km$
(a) $^{238}U$ has 238 nucleons. We can assume that the nucleus is a sphere. We can find the density of the nucleus: $\rho = \frac{m}{V}$ $\rho = \frac{(238)(1.67\times 10^{-27}~kg)}{\frac{4}{3}\pi~r^3}$ $\rho = \frac{(3)(238)(1.67\times 10^{-27}~kg)}{(4\pi)~(7.4\times 10^{-15}~m)^3}$ $\rho = 2.34\times 10^{17}~kg/m^3$ (b) We can find the radius of the neutron star: $\rho = \frac{m}{V}$ $V = \frac{m}{\rho}$ $\frac{4}{3}\pi~r^3 = \frac{m}{\rho}$ $r^3 = \frac{3m}{4\pi~\rho}$ $r = \sqrt[3] {\frac{3m}{4\pi~\rho}}$ $r = \sqrt[3] {\frac{(3)(2.0\times 10^{30}~kg)}{(4\pi)(2.34\times 10^{17}~kg/m^3)}}$ $r = 1.27\times 10^4~m$ $r = 12.7~km$